(05/11/2009, 09:12 PM)BenStandeven Wrote: Actually, , so they are translations of each other, albeit along the imaginary axis instead of the real axis.

Well observed! So this is not even the worst non-uniqueness.
This is generally the case, that there are these two regular super-functions at a fixed point. And one is the other translated by some imaginary value (up to real translations along ) which is half of the period of both functions.

(05/11/2009, 05:17 PM)Ansus Wrote: It should be noted that superfunction is not unique in most cases. For example, for , superfunction is

Ya, this is the simple kind of non-uniqueness, its just a translation along the x-axis.
However there are also more severe types of non-uniques, as I already introduced in my first post, we have two solutions (which are not translations of each other): and .

1. Sorry, Henryk, they are translations of each other..
We already had similar discussion with respect to tetration on base ,http://www.ils.uec.ac.jp/~dima/PAPERS/2009sqrt2.pdf , figure 3.
the growing up SuperExponential (red) and the tetration (blue), at the appropriate translations formula (5.7) and formula (5. become very similar and bounded along the real axis functions (green). (I do not know why the number of forumla that follows (5.7) becomes some strange "smile". I mean just the number of formula, nothing more.)

2. There are many ways to extend the table of superfunctions.
I suggest the group of transforms of the pairs (TransferFunciton, SuperFunctions).

<b>Theorem</b>.
Let ,
Let ,
Let ,
Let .
Then .
Proof:
(end of proof).

With transform , from the pair (f,F) we get the pair (h,E).

2.1. Also, in the right hand side of the expression
we can swap and ;
this gives the new transfer function
with known superfunction .